Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $q = \dfrac{10a - 70}{a^2 - 16a + 63} \div \dfrac{-4a - 28}{a^2 - 9a} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{10a - 70}{a^2 - 16a + 63} \times \dfrac{a^2 - 9a}{-4a - 28} $ First factor the quadratic. $q = \dfrac{10a - 70}{(a - 9)(a - 7)} \times \dfrac{a^2 - 9a}{-4a - 28} $ Then factor out any other terms. $q = \dfrac{10(a - 7)}{(a - 9)(a - 7)} \times \dfrac{a(a - 9)}{-4(a + 7)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ 10(a - 7) \times a(a - 9) } { (a - 9)(a - 7) \times -4(a + 7) } $ $q = \dfrac{ 10a(a - 7)(a - 9)}{ -4(a - 9)(a - 7)(a + 7)} $ Notice that $(a - 7)$ and $(a - 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 10a(a - 7)\cancel{(a - 9)}}{ -4\cancel{(a - 9)}(a - 7)(a + 7)} $ We are dividing by $a - 9$ , so $a - 9 \neq 0$ Therefore, $a \neq 9$ $q = \dfrac{ 10a\cancel{(a - 7)}\cancel{(a - 9)}}{ -4\cancel{(a - 9)}\cancel{(a - 7)}(a + 7)} $ We are dividing by $a - 7$ , so $a - 7 \neq 0$ Therefore, $a \neq 7$ $q = \dfrac{10a}{-4(a + 7)} $ $q = \dfrac{-5a}{2(a + 7)} ; \space a \neq 9 ; \space a \neq 7 $